question_answer 44)

Chlorine is prepared in the laboratory by treating manganese dioxide (\[Mn{{O}_{2}}\]) with aqueous hydrochloric acid according to the reaction: \[4HCl(aq)+Mn{{O}_{2}}(s)\to MnC{{l}_{2}}(aq)\] \[+2{{H}_{2}}O(l)+C{{l}_{2}}(g)\] How many gram of HCl react with 5g of manganese dioxide?

Answer:

The given reaction is, \[\underset{\underset{4\,\,\times \,36.5g}{\mathop{4\,mol}}\,}{\mathop{4HCl(aq)}}\,+\underset{\underset{\text{1 }\!\!\times\!\!\text{ 87g}}{\mathop{\text{1}\,\text{mol}}}\,}{\mathop{Mn{{O}_{2}}(s)}}\,\to \underset{1\,mol}{\mathop{MnC{{l}_{2}}(aq)}}\,+\] \[\underset{2\,mol}{\mathop{2{{H}_{2}}O(l)}}\,+\underset{1\,mol}{\mathop{C{{l}_{2}}(g)}}\,\] \[\because \]\[87gMn{{O}_{2}}\]reacts with \[4\times 36.5g\,HCl\] \[\therefore \]\[5g\,Mn{{O}_{2}}\]will react with \[\frac{4\times 36.5}{87}\times 5g\,HCl\] = 8.39 g